Question: Solve for $k$, $ \dfrac{6}{6k + 6} = \dfrac{k - 7}{3k + 3} - \dfrac{4}{3k + 3} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $6k + 6$ $3k + 3$ and $3k + 3$ The common denominator is $6k + 6$ The denominator of the first term is already $6k + 6$ , so we don't need to change it. To get $6k + 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{k - 7}{3k + 3} \times \dfrac{2}{2} = \dfrac{2k - 14}{6k + 6} $ To get $6k + 6$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{4}{3k + 3} \times \dfrac{2}{2} = -\dfrac{8}{6k + 6} $ This give us: $ \dfrac{6}{6k + 6} = \dfrac{2k - 14}{6k + 6} - \dfrac{8}{6k + 6} $ If we multiply both sides of the equation by $6k + 6$ , we get: $ 6 = 2k - 14 - 8$ $ 6 = 2k - 22$ $ 28 = 2k $ $ k = 14$